Find the equation(s) of the tangent line(s) to the graph of y^2 - xy + 3 = 0 at x = -4. A) y = - 3/2x - 3 B) y = 3/2x + 3 and y = - 1/2x - 3 C) y = 3/2x + 1/2 D) y = 3/2x - 3 Show your work:
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Given equation: \(y^2 + xy + 3 = 0\) Taking the derivative with respect to x: \(2yy' + y + x\frac{dy}{dx} = 0\) Show more…
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